Testing for spatial dependence in a spatial autoregressive (SAR) model in the presence of endogenous regressors

Abstract

Spatial modeling is one of the growing areas of research in economics in recent years. However, these models are not tested enough. Even if tests are performed, they are done in a piece-wise fashion. Another age-long problem in economic modeling is endogeneity of one or more variables. Endgeneity is caused due to a number of reasons one of which is simultaneous modeling of economic variables. This paper considers specification testing in the context of a spatial autoregresive (SAR) model with an endogenous regressor. First, we construct standard Rao’s Score (RS) tests for null hypothesis of the absence of spatial autocorrelation and endogeneity. These standard RS tests are invalid in the presence of local misspecification of the models under the alternative hypotheses. Therefore, in our next step, we develop adjusted tests using the technique of Bera and Yoon (Econom Theor 9:649–658, 1993), that are robust to local misspecification. These adjusted (or robustified) tests are simple to calculate and easy to implement. With a Monte Carlo study we investigate the finite sample performance of all the proposed tests, and the results confirm that the robust tests perform better compared to their non-robust counterparts both in terms of size and power.

Appendices

Appendix 1: Score functions and the Hessian matrix

Using the expression of the log-likelihood function in Eq. (4.1), here we derive the first- and the second-order derivatives.

1.1 Scores

$$\begin{aligned} \left. \begin{aligned} \frac{\partial l(\theta ;y_{1},\,y_{2})}{\partial \phi }&=\frac{y_{2}^{'}\xi }{\sigma _{\xi }^{2}}, \quad \frac{\partial l(\theta ;y_{1},\,y_{2})}{\partial \beta }=\frac{X_{1}^{'}\xi }{\sigma _{\xi }^{2}}, \quad \frac{\partial l(\theta ;y_{1},\,y_{2})}{\partial \gamma }=\frac{X^{'}u_{2}}{\sigma _{2}^{2}}-\frac{\delta X^{'}\xi }{\sigma _{\xi }^2},\\ \frac{\partial l(\theta ;y_{1},\,y_{2})}{\partial \sigma _{\xi }^{2}}&=-\frac{n}{2\sigma _{\xi }^{2}}+\frac{\xi ^{'}\xi }{2\sigma _{\xi }^{4}}, \quad \frac{\partial l(\theta ;y_{1},\,y_{2})}{\partial \sigma _{2 }^{2}}=-\frac{n}{2\sigma _{2 }^{2}}+\frac{u_{2}^{'}u_{2}}{2\sigma _{2 }^{4}}, \quad \\ \frac{\partial l(\theta ;y_{1},\,y_{2})}{\partial \lambda }&=-tr(G)+\frac{y_{1}^{'}W^{'}\xi }{\sigma _{\xi }^{2}}, \quad \frac{\partial l(\theta ;y_{1},\,y_{2})}{\partial \delta }=\frac{u_{2}^{'}\xi }{\sigma _{\xi }^{2}}, \end{aligned} \right\} \end{aligned}$$

where $G=S^{-1}W$ and $S=(I_{n}-\lambda W)$. These score functions evaluated under the joint null $H_{0}:\lambda _{0}=0,\,\delta _{0}=0$, give us the scores in Eq. (4.2) that were used in developing the test statistics in Sect. 4.

Taking derivatives of the above score functions we obtain the second-order derivatives, i.e., the elements of the Hessian matrix, as follows:

1.2 Elements of the Hessian matrix

$$\begin{aligned} \left. \begin{aligned} \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \phi ^{2}}&=-\frac{y_{2}^{'}y_{2}}{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \phi \partial \beta ^{'}}=-\frac{y_{2}^{'}X_{1}}{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \phi \partial \gamma ^{'}} =\frac{\delta y_{2}^{'}X}{\sigma _{\xi }^2},\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \phi \partial \sigma _{\xi }^{2}}&=-\frac{y_{2}^{'}\xi }{\sigma _{\xi }^4}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \phi \partial \lambda } =-\frac{y_{2}^{'}Wy_{1}}{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \phi \partial \delta } =-\frac{y_{2}^{'}u_{2}}{\sigma _{\xi }^2},\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \beta \partial \beta ^{'}}&=-\frac{X_{1}^{'}X_{1}}{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \beta \partial \gamma ^{'}} =\frac{\delta X_{1}^{'}X}{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \beta \partial \sigma _{\xi }^2} = -\frac{X_{1}^{'}\xi }{\sigma _{\xi }^4},\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \beta \partial \lambda }&= -\frac{X_{1}^{'}W y_{1}}{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \beta \partial \delta } = -\frac{X_{1}^{'}u_{2}}{\sigma _{\xi }^2}, \quad\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \gamma \partial \gamma ^{'}} &= - \Big [\frac{X^{'}X}{\sigma _{2}^2} + \frac{\delta ^{2}X^{'}X}{\sigma _{\xi }^2}\Big ],\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \gamma \partial \sigma _{\xi }^2}&= \frac{\delta X^{'}\xi }{\sigma _{\xi }^4}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \gamma \partial \sigma _{2}^{2}} = -\frac{X^{'}u_2}{\sigma _{2}^4}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \gamma \partial \lambda } = \frac{\delta X^{'}Wy_{1}}{\sigma _{\xi }^2},\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \gamma \partial \delta }&= \frac{\delta X^{'}u_{2}-X^{'}\xi }{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial (\sigma _{\xi }^2)^{2}} =\Big [\frac{n}{2\sigma _{\xi }^4} - \frac{\xi ^{'}\xi }{\sigma _{\xi }^6}\Big ],\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \sigma _{\xi }^2\partial \lambda }&= -\frac{y_{1}^{'}W^{'}\xi }{\sigma _{\xi }^4}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \sigma _{\xi }^2\partial \delta }= -\frac{u_{2}^{'}\xi }{\sigma _{\xi }^4}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial (\sigma _{2}^2)^{2}} = \Big [\frac{n}{2\sigma _{2}^4} - \frac{u_{2}^{'}u_{2}}{\sigma _{2}^6}\Big ],\\ \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \lambda ^{2}}&=-tr(G^2) -\frac{y_{1}^{'}W^{'}Wy_{1}}{\sigma _{\xi }^2}, \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \lambda \partial \delta }= -\frac{u_{2}^{'}Wy_{1}}{\sigma _{\xi }^2},\\ \quad \frac{\partial ^{2} l(\theta ; y_{1},\,y_{2})}{\partial \delta ^{2}}&= -\frac{u_{2}^{'}u_{2}}{\sigma _{\xi }^2} \end{aligned} \right\} . \end{aligned}$$

Appendix 2: Proofs of proposition and corollaries

1.1 Proof of Proposition 1

We start with the first result. To do so we derive the asymptotic normality of $d_{\alpha _{2}}(\tilde{\theta })$ under $H_{A}^{\alpha _{2}}:\alpha _{20}=\alpha ^{*}_{2}+\frac{\tau _{2}}{\sqrt{n}}$, and $H_{A}^{\alpha _{3}}:\alpha _{30}=\alpha ^{*}_{3}+\frac{\tau _{3}}{\sqrt{n}}$. Let $\theta =(\alpha _{1}^{'},\,\alpha _{2}^{'},\,\alpha _{3}^{'})^{'}$ be an arbitrary value of $\theta$ and $\tilde{\theta }=(\tilde{\alpha _{1}}^{'},\,\alpha ^{*'}_{2},\,\alpha ^{*'}_{3})^{'}$ be the restricted MLE of $\theta$ under the joint null. Let us expand the score function w.r.t. $\alpha _{2}$, i.e., $\frac{\partial l(\tilde{\theta })}{\partial \alpha _{2}}$, around $\theta _{0}=(\alpha _{10}^{'},\,\alpha _{20}^{'},\,\alpha _{30}^{'})^{'}$ using Taylor series expansion as

$$\begin{aligned} \frac{\partial l(\tilde{\theta })}{\partial \alpha _{2}}\overset{a}{=} \frac{\partial l(\theta _{0})}{\partial \alpha _{2}} + \frac{\partial ^{2} l(\theta _{0})}{\partial \alpha _{2}\partial \alpha _{2}^{'}}(\alpha ^{*}_{2}-\alpha _{20}) + \frac{\partial ^{2} l(\theta _{0})}{\partial \alpha _{2}\partial \alpha _{3}^{'}}(\alpha ^{*}_{3}-\alpha _{30})+\frac{\partial ^{2} l(\theta _{0})}{\partial \alpha _{2}\partial \alpha _{1}^{'}}(\tilde{\alpha _{1}}-\alpha _{10}), \end{aligned}$$

(4.15)

where $``\overset{a}{=}"$ means asymptotic equivalence.

Since $(\alpha ^{*}_{2}-\alpha _{20})=-\frac{\tau _{2}}{\sqrt{n}}$ and $(\alpha ^{*}_{3}-\alpha _{30})=-\frac{\tau _{3}}{\sqrt{n}}$ under $H_{0}$, we have

$$\begin{aligned} \frac{\partial l(\tilde{\theta })}{\partial \alpha _{2}}\overset{a}{=}\frac{\partial l(\theta _{0})}{\partial \alpha _{2}}-\frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{2} \partial \alpha _{2}^{'}}\frac{\tau _{2}}{\sqrt{n}}-\frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{2} \partial \alpha _{3}^{'}}\frac{\tau _{3}}{\sqrt{n}}+\frac{\partial ^{2} l(\theta _{0})}{\partial \alpha _{2}\partial \alpha _{1}^{'}}(\tilde{\alpha _{1}}-\alpha _{10})\nonumber \\ \implies \frac{1}{\sqrt{n}}\frac{\partial l(\tilde{\theta })}{\partial \alpha _{2}}\overset{a}{=}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{2}} + J_{\alpha _{2}\alpha _{2}}\tau _{2} + J_{\alpha _{2}\alpha _{3}}\tau _{3} -J_{\alpha _{2}\alpha _{1}}\sqrt{n}(\tilde{\alpha _{1}}-\alpha _{10}). \end{aligned}$$

(4.16)

Now expanding $\frac{\partial l(\tilde{\theta )}}{\partial \alpha _{1}}$ around $\theta _{0}=(\alpha _{10}^{'},\,\alpha _{20}^{'},\,\alpha _{30}^{'})^{'}$, we obtain

$$\begin{aligned} \frac{\partial l(\tilde{\theta )}}{\partial \alpha _{1}}&\overset{a}{=}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}} + \frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{2}^{'}}(\alpha ^{*}_{2}-\alpha _{20}) + \frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{3}^{'}}(\alpha ^{*}_{3}-\alpha _{30}) + \frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{1}^{'}}(\tilde{\alpha _{1}}-\alpha _{10}). \end{aligned}$$

Since $\frac{\partial l(\tilde{\theta )}}{\partial \alpha _{1}}=0$, where $\tilde{\theta }=(\tilde{\alpha _{1}}^{'},\,\alpha ^{*'}_{2},\,\alpha ^{*'}_{3})^{'}$ is the restricted MLE of $\theta$, we have from the above equation

$$\begin{aligned} 0&\overset{a}{=}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}} + \frac{1}{\sqrt{n}}\frac{\partial l^{2}(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{2}^{'}}(\alpha ^{*}_{2}-\alpha _{20}) + \frac{1}{\sqrt{n}}\frac{\partial l^{2}(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{3}^{'}}(\alpha ^{*}_{3}-\alpha _{30}) + \frac{1}{\sqrt{n}}\frac{\partial l^{2}(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{1}^{'}}(\tilde{\alpha _{1}}-\alpha _{10})\nonumber \\&\implies \frac{1}{n}\frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{1}^{'}}\sqrt{n}(\tilde{\alpha _{1}}-\alpha _{10})\overset{a}{=}-\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}} + \frac{1}{n}\frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{2}}\tau _{2} + \frac{1}{n}\frac{\partial ^{2} l(\theta _{0})}{\partial \alpha _{1} \partial \alpha _{3}^{'}}\tau _{3}\nonumber \\&\implies -J_{\alpha _{1}\alpha _{1}}\sqrt{n}(\tilde{\alpha _{1}}-\alpha _{10})\overset{a}{=}-\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}}-J_{\alpha _{1}\alpha _{2}}\tau _{2} -J_{\alpha _{1}\alpha _{3}}\tau _{3}\nonumber \\&\implies \sqrt{n}(\tilde{\alpha _{1}}-\alpha _{10})\overset{a}{=}J^{-1}_{\alpha _{1}\alpha _{1}}\Big (\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}} + J_{\alpha _{1}\alpha _{2}}\tau _{2} + J_{\alpha _{1}\alpha _{3}}\tau _{3}\Big ). \end{aligned}$$

(4.17)

Substituting the expression of $\sqrt{n}(\tilde{\alpha _{1}}-\alpha _{10})$ from (4.17) in Eq. (4.16) we obtain

$$\begin{aligned} \frac{1}{\sqrt{n}}\frac{\partial l(\tilde{\theta })}{\partial \alpha _{2}}&\overset{a}{=}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{2}} + J_{\alpha _{2}\alpha _{2}}\tau _{2} + J_{\alpha _{2}\alpha _{3}}\tau _{3} - J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}[\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}} + J_{\alpha _{1}\alpha _{2}}\tau _{2} + J_{\alpha _{1}\alpha _{3}}\tau _{3}]\\&\overset{a}{=}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{2}} + J_{\alpha _{2}\alpha _{2}}\tau _{2} + J_{\alpha _{2}\alpha _{3}}\tau _{3} -J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}J_{\alpha _{1}\alpha _{2}}\tau _{2}\\&- J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}J_{\alpha _{1}\alpha _{3}}\tau _{3}\\&\overset{a}{=}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{2}} + (J_{\alpha _{2}\alpha _{2}}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}J_{\alpha _{1}\alpha _{2}})\tau _{2}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}}\\&+ (J_{\alpha _{2}\alpha _{3}}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}J_{\alpha _{1}\alpha _{3}})\tau _{3}. \end{aligned}$$

Denoting $J_{\alpha _{2}\cdot \alpha _{1}}\equiv J_{\alpha _{2}\alpha _{2}} - J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}J_{\alpha _{1}\alpha _{2}}$ and $J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}} \equiv J_{\alpha _{2}\alpha _{3}}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}J_{\alpha _{1}\alpha _{2}}$, from the above equation we have

$$\begin{aligned} \frac{1}{\sqrt{n}}\frac{\partial l(\tilde{\theta })}{\partial \alpha _{2}}&\overset{a}{=} [\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{2}}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}}] + [J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2} + J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3}]\\&=R + C ,\, \text {say}, \end{aligned}$$

where $R=\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{2}}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}\frac{1}{\sqrt{n}}\frac{\partial l(\theta _{0})}{\partial \alpha _{1}}$ is a random variable with mean zero while $C=J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2} + J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3}$ is simply a constant.

Therefore,

$$\begin{aligned} \sqrt{n}d_{\alpha _{2}}(\tilde{\theta })&\overset{d}{\rightarrow }N(C,\,V), \end{aligned}$$

where $V=Var(R)=(J_{\alpha _{2}\alpha _{2}}-J_{\alpha _{2}\alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}}J_{\alpha _{1}\alpha _{2}})=J_{\alpha _{2}\cdot \alpha _{1}}$. Thus, we can write

$$\begin{aligned} \sqrt{n}d_{\alpha _{2}}(\tilde{\theta })&\overset{d}{\rightarrow } N(J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2} + J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3},\,\,J_{\alpha _{2}\cdot \alpha _{1}}). \end{aligned}$$

(4.18)

The asymptotic distribution of $RS_{\alpha _{2}}$ can be easily obtained using the asymptotic distribution of $d_{\alpha _{2}}$ in Eq. (4.18) as follows:

$$\begin{aligned} RS_{\alpha _{2}}&=nd^{'}_{\alpha _{2}}(\tilde{\theta })J^{-1}_{\alpha _{2}\cdot \alpha _{1}}(\tilde{\theta })d_{\alpha _{2}}(\tilde{\theta })\overset{d}{\rightarrow }\chi ^{2}_{r}(\nu _{3}), \end{aligned}$$

(4.19)

where $\nu _{3}\equiv \nu _{3}(\tau _{2},\,\tau _{3})=(J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2} + J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3})^{'}J^{-1}_{\alpha _{2}\cdot \alpha _{1}}(J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2} + J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3})=\tau _{2}^{'}J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2} + 2\tau _{2}^{'}J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3} + \tau _{3}^{'}J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}J^{-1}_{\alpha _{2}\cdot \alpha _{1}}J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3}$.

This proves the first part of Proposition 1. The second part of the proposition states:

Under $H_{o}^{\alpha _{2}}:\alpha _{20}=\alpha ^{*}_{2}$ and irrespective of the value of $\alpha _{3}$ we have

$$\begin{aligned} RS^{*}_{\alpha _{2}}\overset{d}{\rightarrow }\chi ^{2}_{q}. \end{aligned}$$

To do this let us define $\kappa =(\alpha _{2}^{'},\,\alpha _{3}^{'})^{'}$ and $l_{\kappa }(\tilde{\theta })=\frac{\partial l(\tilde{\theta )}}{\partial \kappa }=\big [l^{'}_{\alpha _{2}}(\tilde{\theta }),\,l^{'}_{\alpha _{3}}(\tilde{\theta })\big ]^{'}$, say.

Expanding $l_{\kappa }(\tilde{\theta })$ around $\theta _{0}=(\alpha ^{'}_{10},\,\kappa ^{'}_{0})^{'}$ using first-order Taylor series expansion, we get

$$\begin{aligned} \frac{1}{\sqrt{n}} l_{\kappa }(\tilde{\theta })&\overset{a}{=}\frac{1}{\sqrt{n}}l_{\kappa }(\theta _{0})+\frac{1}{\sqrt{n}}\frac{\partial ^{2} l(\theta _{0})}{\partial \kappa \partial \kappa ^{'}}(\tilde{\kappa }-\kappa _{0})+ \frac{1}{\sqrt{n}}\frac{\partial ^{2}l(\theta _{0}}{\partial \kappa \partial \alpha _{1}}(\tilde{\alpha _{1}}-\alpha _{10})\nonumber \\ \implies \sqrt{n}d_{\kappa }(\tilde{\theta })&\overset{a}{=}\sqrt{n}d_{\kappa }(\theta _{0})-\sqrt{n}D_{\kappa \kappa }(\theta _{0})(\tau _{2}^{'},\,\tau _{3}^{'})^{'} + \sqrt{n}D_{\kappa \alpha _{1}}(\theta _{0})(\tilde{\alpha _{1}}-\alpha _{10}), \end{aligned}$$

(4.20)

where D denotes the second order derivative matrix, namely

$$\begin{aligned} D_{\kappa \alpha _{1}}=\begin{pmatrix} d_{\alpha _{2}\alpha _{1}}\\ d_{\alpha _{3}\alpha _{1}} \end{pmatrix} \quad \text {and}\quad D_{\kappa \kappa }=\begin{pmatrix} d_{\alpha _{2}\alpha _{2}} &{} d_{\alpha _{2}\alpha _{3}}\\ d_{\alpha _{3}\alpha _{2}} &{} d_{\alpha _{3}\alpha _{3}} \end{pmatrix}. \end{aligned}$$

Similarly, expanding $l_{\alpha _{1}}(\tilde{\theta })$ by Taylor series expansion we obtain

$$\begin{aligned} \frac{1}{\sqrt{n}}l_{\alpha _{1}}(\tilde{\theta })&\overset{a}{=}\frac{1}{\sqrt{n}}l_{\alpha _{1}}(\theta _{0}) + \frac{1}{\sqrt{n}}\frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{1} \partial \kappa ^{'}}(\tilde{\kappa }-\kappa _{0}) + \frac{1}{\sqrt{n}}\frac{\partial ^{2}l(\theta _{0})}{\partial \alpha _{1} \partial \alpha ^{'}_{1}}(\tilde{\alpha _{1}}-\alpha _{10})\nonumber \\ \implies \sqrt{n}d_{\alpha _{1}}(\tilde{\theta })&\overset{a}{=}\sqrt{n}d_{\alpha _{1}}(\theta _{0}) - D_{\alpha _{1}\kappa }(\theta _{0})(\tau _{2}^{'},\,\tau _{3}^{'})^{'} +\sqrt{n}D_{\alpha _{1}\alpha _{1}}(\theta _{0})(\tilde{\alpha _{1}}-\alpha _{1}). \end{aligned}$$

(4.21)

Combining Eqs. (4.20) and (4.21), we get

$$\begin{aligned} \sqrt{n}d_{\kappa }(\tilde{\theta })&\overset{a}{=}(-J_{\kappa \alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}},\,I_{q+r})\begin{pmatrix} \frac{1}{\sqrt{n}}d_{\alpha _{1}}(\theta _{0})\\ \frac{1}{\sqrt{n}}d_{\kappa }(\theta _{0})\end{pmatrix} + \begin{pmatrix} J_{\alpha _{2}\cdot \alpha _{1}} &{} J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\\ J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}} &{} J_{\alpha _{3}\cdot \alpha _{1}}\end{pmatrix}\begin{pmatrix} \tau _{2}\\ \tau _{3} \end{pmatrix}, \end{aligned}$$

(4.22)

where $J_{\kappa \alpha _{1}}=\begin{pmatrix} J_{\alpha _{2}\alpha _{1}}\\ J_{\alpha _{3}\alpha _{1}} \end{pmatrix}$. Substituting $\tau _{2}=0$ in Eq. (4.22) gives us distribution of $\sqrt{n}d_{\kappa }(\tilde{\theta })$ under $H^{\alpha _{2}}_{0}:\alpha _{20}=\alpha ^{*}_{2}$ and $H^{\alpha _{3}}_{A}:\alpha _{30}=\alpha ^{*}_{3}+\tau _{3}/\sqrt{n}$ as

$$\begin{aligned} \sqrt{n}d_{\kappa }(\tilde{\theta })\overset{a}{=}(-J_{\kappa \alpha _{1}}J^{-1}_{\alpha _{1}\alpha _{1}},\,I_{q+r})\begin{pmatrix} \frac{1}{\sqrt{n}}d_{\alpha _{1}}(\theta _{0})\\ \frac{1}{\sqrt{n}}d_{\kappa }(\theta _{0}) \end{pmatrix} + \begin{pmatrix} J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3} \\ J_{\alpha _{3}\cdot \alpha _{1}} \tau _{3} \end{pmatrix} \end{aligned}$$

implying

$$\begin{aligned} \sqrt{n}d_{\kappa }(\tilde{\theta })\overset{d}{\rightarrow }N\Big [\begin{pmatrix} J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\tau _{3}\\ J_{\alpha _{3}\cdot \alpha _{1}}\tau _{3} \end{pmatrix},\,\begin{pmatrix} J_{\alpha _{2}\cdot \alpha _{1}} &{} J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\\ J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}} &{} J_{\alpha _{3}\cdot \alpha _{1}} \end{pmatrix}\Big ]. \end{aligned}$$

(4.23)

The adjusted score w.r.t. $\alpha _{2}$ can be written as

$$\begin{aligned} \sqrt{n} d^{*}_{\alpha _{2}}(\tilde{\theta })&=(I_{q},\,-J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}J^{-1}_{\alpha _{3}\cdot \alpha _{1}})d_{\kappa }(\tilde{\theta }). \end{aligned}$$

The distribution of the adjusted score is obtained from the distribution in Eq. (4.23) as

$$\begin{aligned} \sqrt{n}d^{*}_{\alpha _{2}}(\tilde{\theta })\overset{d}{\rightarrow }N\big [0,\,J_{\alpha _{2}\cdot \alpha _{1}}-J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}J^{-1}_{\alpha _{3}\cdot \alpha _{1}}J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}}\big ]. \end{aligned}$$

Thus, for $RS^{*}_{\alpha _{2}}$ which is based on $d^{*}_{\alpha _{2}}$, we have the following distribution under $H^{\alpha _{2}}_{0}:\alpha _{20}=\alpha ^{*}_{2}$ and irrespective of the value of $\alpha _{3}$

$$\begin{aligned} RS^{*}_{\alpha _{2}}\overset{d}{\rightarrow }\chi ^{2}_{q}. \end{aligned}$$

Finally, the last part of Proposition 1 states:

Under $H^{\alpha _{2}}_{A}:\alpha _{20}=\alpha ^{*}_{2}+\tau _{2}/\sqrt{n}$ and irrespective of the value of $\alpha _{3}$

$$\begin{aligned} RS^{*}_{\alpha _{2}}\overset{d}{\rightarrow }\chi ^{2}_{q}(\nu _{4}), \end{aligned}$$

where $\nu _{4}\equiv \nu _{4}(\tau _{2})=\tau _{2}^{'}J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2}-\tau _{2}^{'}J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}J^{-1}_{\alpha _{3}\cdot \alpha _{1}}J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}}\tau _{2}$ is the non-centrality parameter. To prove this, note that similarly as in (4.23), the distribution of $\sqrt{n}d_{\kappa }(\tilde{\theta })$ under $H^{\alpha _{2}}_{A}$ and $H^{\alpha _{3}}_{0}$ can be obtained by substituting $\tau _{3}=0$ in (4.22) as

$$\begin{aligned} \sqrt{n}d_{\kappa }(\tilde{\theta })\overset{d}{\rightarrow }N\Big [\begin{pmatrix} J_{\alpha _{2}\cdot \alpha _{1}}\tau _{2}\\ J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}}\tau _{2} \end{pmatrix},\,\begin{pmatrix} J_{\alpha _{2}\cdot \alpha _{1}} &{} J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}\\ J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}} &{} J_{\alpha _{3}\cdot \alpha _{1}} \end{pmatrix}\Big ]. \end{aligned}$$

(4.24)

Thus, the distribution of $d^{*}_{\alpha _{2}}$ can be obtained from (4.24) as

$$\begin{aligned} \sqrt{n}d^{*}_{\alpha _{2}}(\tilde{\theta })&\overset{d}{\rightarrow }N\big [(J_{\alpha _{2}\cdot \alpha _{1}}-J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}J^{-1}_{\alpha _{3}\cdot \alpha _{1}}J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}})\tau _{2},\,J_{\alpha _{2}\cdot \alpha _{1}}-J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}J^{-1}_{\alpha _{3}\cdot \alpha _{1}}J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}}\big ], \end{aligned}$$

which implies

$$\begin{aligned} RS^{*}_{\alpha _{2}}\overset{d}{\rightarrow }\chi ^{2}_{r}(\nu _{4}), \end{aligned}$$

where $\nu _{4}\equiv \nu _{4}(\tau _{2})=\tau _{2}^{'}(J_{\alpha _{2}\cdot \alpha _{1}}-J_{\alpha _{2}\alpha _{3}\cdot \alpha _{1}}J^{-1}_{\alpha _{3}\cdot \alpha _{1}}J_{\alpha _{3}\alpha _{2}\cdot \alpha _{1}})\tau _{2}$.

1.2 Proof of Corollaries

In this section we derive the non centrality parameters of the tests in the context of our model as given in Corollaries 2 and 3.

Proof of Corollary 2

The first part of Corollary 2 states that

Under $H^{\lambda }_{A}:\lambda _{0}=\frac{\tau _{2}}{\sqrt{n}}$ and $H^{\delta }_{A}:\delta _{0}=\frac{\tau _{3}}{\sqrt{n}}$

$$\begin{aligned} RS_{\lambda }\overset{d}{\rightarrow } \chi ^{2}_{1}(\nu _{9}), \end{aligned}$$

where $\nu _{9}\equiv \nu _{9}(\tau _{2},\,\tau _{3})=\tau ^{2}_{2}J_{\lambda \cdot \psi } + 2\tau _{2}\tau _{3}J_{\lambda \delta \cdot \psi } + \tau ^{2}_{3}\frac{J^{2}_{\lambda \delta \cdot \psi }}{J_{\lambda \cdot \psi }}$. Comparing with our general set-up in Sect. 3, here we have $\alpha _{2}\equiv \lambda$, $\alpha _{3}\equiv \delta$ and the set of nuisance parameters as $\alpha _{1}\equiv \psi =(\phi ,\,\beta ^{'},\,\gamma ^{'},\,\sigma _{\xi }^{2},\,\sigma _{2}^{2})^{'}$. Therefore, substituting these in $\nu _{3}$ of Proposition 1 we get

$$\begin{aligned} \nu _{9}&=\tau ^{'}_{2}J_{\lambda \cdot \psi }\tau _{2} + 2\tau ^{'}_{2}J_{\lambda \delta \cdot \psi }\tau _{3} + \tau ^{'}_{3}J^{'}_{\lambda \delta \cdot \psi }J^{-1}_{\lambda \cdot \psi }J_{\lambda \delta \cdot \psi }\tau _{3}\\&=\tau ^{2}_{2}J_{\lambda \cdot \psi } + 2\tau _{2}\tau _{3}J_{\lambda \delta \cdot \psi } + \tau ^{2}_{3}\frac{J^{2}_{\lambda \delta \cdot \psi }}{J_{\lambda \cdot \psi }}, \end{aligned}$$

The second part of Corollary 2 states

Under $H^{\lambda }_{A}:\lambda _{0}=\frac{\tau _{2}}{\sqrt{n}}$

$$\begin{aligned} RS^{*}_{\lambda }\overset{d}{\rightarrow }\chi ^{2}_{1}(\nu _{10}), \end{aligned}$$

where $\nu _{10}\equiv \nu _{10}(\tau _{2})=\tau ^{2}_{2}[J_{\lambda \cdot \psi }-\frac{J^{2}_{\lambda \delta \cdot \psi }}{J_{\delta \cdot \psi }}]$.

The proof for the non-centrality parameter directly follows from $\nu _{4}$ in the last part of Proposition 1.

$$\begin{aligned} \nu _{10}&=\tau ^{'}_{2}J_{\lambda \cdot \psi }\tau _{2} - \tau ^{'}_{2}J_{\lambda \delta \cdot \psi }J^{-1}_{\delta \cdot \psi }J_{\lambda \delta \cdot \psi }\tau _{2} \\&=\tau ^{2}_{2}\big [J_{\lambda \cdot \psi }-\frac{J^{2}_{\lambda \delta \cdot \psi }}{J_{\delta \cdot \psi }}\big ]. \end{aligned}$$

$\square$

Proof of Corollary 3

Similarly as in Corollary 2, we obtain the non centrality parameters in Corollary 3 by substituting $\alpha _{2}\equiv \delta$ and $\alpha _{3}\equiv \lambda$ and $\alpha _{1}\equiv \psi =(\phi ,\,\beta ^{'},\,\gamma ^{'},\,\sigma _{\xi }^{2},\,\sigma _{2}^{2})^{'}$.

The first part of Corollary 3 says

Under $H^{\delta }_{A}:\delta _{0}=\frac{\tau _{3}}{\sqrt{n}}$ and $H^{\lambda }_{A}:\lambda _{0}=\frac{\tau _{2}}{\sqrt{n}}$

$$\begin{aligned} RS_{\delta }\overset{d}{\rightarrow } \chi ^{2}_{1}(\nu _{11}), \end{aligned}$$

where $\nu _{11}\equiv \nu _{11}(\tau _{2},\,\tau _{3})=\tau ^{2}_{3}J_{\delta \cdot \psi } + 2\tau _{2}\tau _{3}J_{\lambda \delta \cdot \psi } + \tau ^{2}_{2}\frac{J^{2}_{\lambda \delta \cdot \psi }}{J_{\delta \cdot \psi }}$.

Following Proposition 1 we have

$$\begin{aligned} \nu _{11}&=\tau ^{'}_{3}J_{\delta \cdot \psi }\tau _{3} + 2\tau ^{'}_{3}J_{\delta \delta \cdot \psi }\tau _{2} + \tau ^{'}_{2}J^{'}_{\delta \lambda \cdot \psi }J^{-1}_{\delta \cdot \psi }J_{\delta \lambda \cdot \psi }\tau _{2}\\&=\tau ^{3}_{2}J_{\delta \cdot \psi } + 2\tau _{3}\tau _{2}J_{\delta \lambda \cdot \psi } + \tau ^{2}_{2}\frac{J^{2}_{\delta \lambda \cdot \psi }}{J_{\delta \cdot \psi }}. \end{aligned}$$

The second part of Corollary 3 states

Under $H^{\delta }_{A}:\delta =\frac{\tau _{3}}{\sqrt{n}}$ we have

$$\begin{aligned} RS^{*}_{\delta }\overset{d}{\rightarrow }\chi ^{2}_{1}(\nu _{12}), \end{aligned}$$

where $\nu _{12}\equiv \nu _{12}(\tau _{3})=\tau ^{2}_{3}[J_{\delta \cdot \psi }-\frac{J^{2}_{\lambda \delta \cdot \psi }}{J_{\lambda \cdot \psi }}]$.

Again from Proposition 1 we directly obtain

$$\begin{aligned} \nu _{12}&=\tau ^{'}_{3}J_{\delta \cdot \psi }\tau _{3} - \tau ^{'}_{3}J_{\delta \lambda \cdot \psi }J^{-1}_{\lambda \cdot \psi }J_{\delta \lambda \cdot \psi }\tau _{3} \\&=\tau ^{2}_{3}\big [J_{\delta \cdot \psi }-\frac{J^{2}_{\delta \lambda \cdot \psi }}{J_{\lambda \cdot \psi }}\big ]. \end{aligned}$$

$\square$